**Math Calculators** ▶ Fourier Series Calculator

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**Table of Content**

An online Fourier series calculator is exclusively designed to calculate the Fourier series of the given periodic function. Now, we have decided to commence with some basic theory!

Read on now!

In mathematics,

“**The expansion of the periodic function in terms of infinite sums of sines and cosines is known as Fourier series.”**

You can calculate the expansion of the function with the help of free online Fourier series calculator.

Take a look at the given formula that shows the periodic function f(x) in the interval \(-L\le \:x\le \:L\:\)

$$ f\left(x\right)=a_0+\sum _{n=1}^{\infty \:}a_n\cdot \cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty \:}b_n\cdot \sin \left(\frac{n\pi x}{L}\right) $$

where ;

$$ a_0=\frac{1}{2L}\cdot \int _{-L}^Lf\left(x\right)dx $$

$$ a_n=\frac{1}{L}\cdot \int _{-L}^Lf\left(x\right)\cos \left(\frac{n\pi x}{L}\right)dx,\:\quad \:n>0 $$

$$ b_n=\frac{1}{L}\cdot \int _{-L}^Lf\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx,\:\quad \:n>0 $$

With the help of the Fourier coefficients calculator, you can easily find values against these coefficients.

In some of the problems that we face regarding Fourier series, the Fourier coefficients **a_{0}**, **a_{n}** or** b_{n}** may become zero after the integration is done. It means that if we try to find the zero coefficients, it may be a time-consuming process indeed and should be avoided. But if we have proper knowledge of even and odd functions, it becomes easy for us to predict the zero coefficients before we start integration.

**Even Functions:**

A function f(x) is said to be an even function if it satisfied the following condition:

**f (-x) = f (x)**

For more details, click Even or Odd Function.

For an even function f(x) that is defined over the period **-L, L** and having a period of **2L**, the coefficient **b_{n}** becomes zero.

$$ b_n=\frac{1}{L}\cdot \int _{-L}^Lf\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx,\:\quad \:n>0 $$

$$ b_n = 0 $$

So for an even function, we only need to determine the coefficients** a_{0}** and **a_{n}** as follows:

$$ a_0=\frac{1}{2L}\cdot \int _{-L}^Lf\left(x\right)dx $$

$$ a_n=\frac{1}{L}\cdot \int _{-L}^Lf\left(x\right)\cos \left(\frac{n\pi x}{L}\right)dx,\:\quad \:n>0 $$

So, for an even function, the Fourier expansion only contains the cosine terms.

$$ f\left(x\right)=a_0+\sum _{n=1}^{\infty \:}a_n\cdot \cos \left(\frac{n\pi x}{L}\right) $$

Whenever you come across an even function, you may use our free online Fourier cosine series calculator.

**Odd Functions:**

In general, a function f(x) is said to be an odd function if;

**f(-x) = -f(x)**

For more details, click Even or Odd Function.

If an odd function is defined over the period **-L, L** and have a time period of **2L**, then we can say that the coefficients **a_{0}** and **a_{n}** becomes zero.

$$ a_n=\frac{1}{L}\cdot \int _{-L}^Lf\left(x\right)\cos \left(\frac{n\pi x}{L}\right)dx,\:\quad \:n>0 $$

$$ a_n = 0 $$

For an odd function given, only one fourier coefficient needs to be determined which are as follows:

$$ b_n=\frac{1}{L}\cdot \int _{-L}^Lf\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx,\:\quad \:n>0 $$

So, for an odd function, the Fourier expansion is only the sine term.

$$ f\left(x\right)=\sum _{n=1}^{\infty \:}b_n\cdot \sin \left(\frac{n\pi x}{L}\right) $$

A fourier sine series calculator is the best way to find the fourier series of an odd function given.

**Properties of Even & Odd Function:**

While dealing with the Fourier series, we must have a proper idea about the basic stuff of even and odd functions that includes:

**Addition Properties:**

- The sum of two even functions is always an even function
- The sum of two odd functions is also odd.
- The sum of an even function and an odd function is neither even nor odd(unless one function is zero)

**Multiplication Properties:**

- The product of two odd functions is aways even
- The product of two even functions is even
- If we multiply an even function with an odd function, we will get an odd function.

A free Fourier series expansion calculator automatically predicts the type of function and calculates the coefficients required to obtain the Fourier series.

To determine the Fourier series of a function given may be a hectic and lengthy practice. That is why we have programmed our free online Fourier series calculator to determine the results instantly and precisely. But to understand the proper usage of Fourier series, let us solve a couple of examples.

**Example # 01:**

Calculate fourier series of the function given below:

$$ f\left( x \right) = L – x on – L \le x \le L $$

**Solution:**

As,

$$ f\left( x \right) = L – x $$

$$ f\left( -x \right) = -(L – x) $$

$$ f\left( x \right) = -f\left( x \right) $$

The given function is odd.

Now, determining the coefficients as follows:

$$ {a_0} = \frac{1}{{2L}}\int_{{\, – L}}^{{\,L}}{{f\left( x \right)\,dx}} $$

$$ {a_0} = \frac{1}{{2L}}\int_{{\, – L}}^{{\,L}}{{L – x\,dx}} $$

$$ {a_0} = 2L $$Â Â (click integral calculator for step by step calculations)

As we know that for an odd function, a_{n} is 0.

Determining the value of b_{n} as follows:

$$ \begin{align*}{B_{\,n}} &= \frac{1}{L}\int_{{\, – L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{1}{L}\int_{{\, – L}}^{{\,L}}{{\left( {L – x} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\\ & = \frac{1}{L}\left. {\left( { – \frac{L}{{{n^2}{\pi ^2}}}} \right)\left[ {L\sin \left( {\frac{{n\,\pi x}}{L}} \right) – n\pi \left( {x – L} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right]} \right|_{ – L}^L\\ & = \frac{1}{L}\left[ {\frac{{{L^2}}}{{{n^2}{\pi ^2}}}\left( {2n\pi \cos \left( {n\pi } \right) – 2\sin \left( {n\pi } \right)} \right)} \right] = \frac{{2L{{\left( { – 1} \right)}^n}}}{{n\pi }}\hspace{0.25in}\hspace{0.25in}n = 1,2,3, \ldots \end{align*} $$Â (click integral calculator for step by step calculations)

In this case, **a_{0}** is not zero but **a_{n}** is **0.**

So, the fourier series is given as:

$$ f\left(x\right)=2 L +\sum _{n=1}^{\infty \:}0\cdot \cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty \:}\frac{2 \left(-1\right)^{n}}{n}\cdot \sin \left(\frac{n\pi x}{L}\right) $$

$$ f\left(x\right)=L + \sum_{n=1}^{\infty} \frac{2 \left(-1\right)^{n} \sin{\left(n x \right)}}{n} $$

Even here, a fourier coefficient calculator helps you to do particular calculations.

**Example # 02:**

Solve fourier series for the function given below:

$$ f\left( x \right) = x^{2} $$

**Solution:**

As we have:

$$ f\left( x \right) = x^{2} $$

$$ f\left( -x \right) = (-x)^{2} $$

$$ f\left( -x \right) = x^{2} $$

$$ f\left( -x \right) = f\left( x \right) $$

So, the given function is an even function for which the value of the Fourier coefficient **b_{n}** becomes zero. The rest of the two coefficients are calculated as follows:

$$ a_0=\frac{1}{2L}\cdot \int _{-L}^Lf\left(x\right)dx $$

$$ a_0=\frac{1}{2\times \pi}\cdot \int _{- \pi}^\pi\left(x^{2}\right)dx $$

$$ a_0= \frac{2 \pi^{2}}{3} $$Â Â (click integral calculator for step by step calculations)

Now, we have:

$$ a_n=\frac{1}{\pi}\cdot \int _{- \pi}^\pi\left(x^{2}\right)\cos \left(\frac{n\pi x}{\pi}\right)dx $$

$$ a_n= \frac{4 \left(-1\right)^{n}}{n^{2}}$$Â Â (click integral calculator for step by step calculations)

The fourier series for the function given is as follows:

$$ f\left(x\right)=a_0+\sum _{n=1}^{\infty \:}a_n\cdot \cos \left(\frac{n\pi x}{L}\right) $$

After putting the values of the coefficients, we get:

$$ f\left(x\right)=\frac{2 \pi^{2}}{3} +\sum _{n=1}^{\infty \:}\frac{4 \left(-1\right)^{n}}{n^{2}}\cdot \cos \left(\frac{n\pi x}{L}\right) $$

$$ \sum_{n=1}^{\infty} \frac{4 \left(-1\right)^{n} \cos{\left(n x \right)}}{n^{2}} + \frac{\pi^{2}}{3} $$

However, for more accurate results, use our free online Fourier series coefficients calculator.

Whenever you come across complex functions, our free online fourier series calculator is here to help you out in determining accurate results. You will get a proper scenario of the calculations by using our calculator.

Let us see what you need to do:

**Input:**

- First, write your function in the drop down list
- After this, select the variable w.r t which you need to determine the Fourier series expansion
- Input the lower and upper limits
- Click â€˜calculateâ€™

**Output:**

The Fourier expansion calculator calculates:

- Fourier series of the function given
- Fourier coefficients of the function
**f: a_{0}, a_{n}, and b_{n}** - Step by step calculations involved in the process

Yes, for any number n, the Fourier series is always unique.

Fourier series makes use of the orthogonality relationship- among sines and cosine functions.

If we have any function that is defined over the entire real line and is periodic is considered to have the Fourier series. However, our free online fourier series calculator helps you to identify whether the given function contains fourier series or not.

The primary use of the Fourier series is that we can easily analyze a signal in another suitable domain rather than its original domain.

Fourier series have very broad prospects in the field of engineering and technologies because it is very crucial in signal processing, acoustics, shell theory, quantum mechanics, and image processing where a vast use of free online fourier series calculator is carried out. Not only this, but you can also resolve complex problems easily by changing the domains of the signals. Moreover, the main advantage of Fourier analysis is that during the process of signal transformation, very little information is lost.

From the source of Wikipedia: Convergence, Fourier series on a square, Hilbert space interpretation, Properties, Riemannâ€“Lebesgue lemma, Riemannâ€“Lebesgue lemma.

From the source of khan academy: Fourier coefficients, Integral of sin(mt) and cos(mt), Integral of product of sines.

From the source of lumen learning: Sine Wave, Arbitrary Wave, Wave Equation.